What Completeness Adds to the Cauchy Property
The Cauchy Criterion for Real Sequences says that a real sequence converges to a finite real limit if and only if it is Cauchy. The word “real” is essential. A sequence of rational numbers is also a real sequence, so it has a real limit whenever it is Cauchy; but that limit need not be rational. Completeness of a set is exactly the guarantee that Cauchy sequences whose terms stay in the set have their limits there as well.
The previous tutorial characterized complete subsets of \(\mathbb{R}\): a nonempty subset is complete if and only if it is closed in \(\mathbb{R}\). We now examine why that condition matters in practice. We will see that without completeness, even a bounded increasing sequence may fail to converge within the set, and a bounded set may have no least upper bound in the set.
A Cauchy Sequence of Rationals Can Have an Irrational Limit
Consider decimal truncations of \(\sqrt{3}\). For \(n\in\mathbb{N}_0\), define $$ q_n=\frac{\lfloor 10^n\sqrt{3}\rfloor}{10^n}. $$ Every \(q_n\) is rational. The defining property of the floor gives $$ 0\leq \sqrt{3}-q_n<\frac{1}{10^n}. $$ This estimate shows that the rational terms get arbitrarily close to \(\sqrt{3}\), even though \(\sqrt{3}\) is not rational.
Worked Example: Rational Decimal Truncations of \(\sqrt{3}\)
The first terms illustrate how the construction works: \(q_0=1\), \(q_1=1.7\), and \(q_2=1.73\). For instance, \(q_1=\lfloor 10\sqrt{3}\rfloor/10=17/10\), since \(17<10\sqrt{3}<18\); and \(q_2=\lfloor 100\sqrt{3}\rfloor/100=173/100\), since \(173<100\sqrt{3}<174\). Each term is rational, while the error estimate is $$ 0\leq\sqrt{3}-q_n<10^{-n}. $$ If \(m,n\geq N\), the triangle inequality gives $$ |q_m-q_n| \leq |q_m-\sqrt{3}|+|q_n-\sqrt{3}| <10^{-m}+10^{-n} \leq 2\cdot 10^{-N}. $$ Given \(\varepsilon>0\), choose \(N\) so that \(2\cdot10^{-N}<\varepsilon\). Then \(|q_m-q_n|<\varepsilon\) whenever \(m,n\geq N\), so \((q_n)\) is Cauchy.
The error estimate also proves \(q_n\to\sqrt{3}\) as a real sequence. But \(\sqrt{3}\notin\mathbb{Q}\). To verify this, suppose \(\sqrt{3}=p/r\) for relatively prime positive integers \(p,r\). Then \(p^2=3r^2\), so \(3\) divides \(p^2\), and hence \(3\) divides \(p\). Write \(p=3k\). Substitution gives \(9k^2=3r^2\), so \(r^2=3k^2\) and \(3\) divides \(r\), contradicting that \(p\) and \(r\) are relatively prime.
If \((q_n)\) converged to a rational number, it would have two real limits: that rational number and \(\sqrt{3}\). The Theorem of Uniqueness of Limits would make them equal, which is impossible. Thus \((q_n)\) is Cauchy as a sequence of rationals but does not converge to a rational number. This is a concrete failure of completeness in \(\mathbb{Q}\).
The example does not contradict the Cauchy Criterion for Real Sequences. The sequence converges in \(\mathbb{R}\), precisely as that criterion predicts. What fails is the stronger assertion that its limit belongs to \(\mathbb{Q}\). This distinction between convergence in an ambient space and convergence within a subset is the point of completeness.
Why Bounded Monotone Sequences Also Need Completeness
The Monotone Convergence Theorem says that every nondecreasing real sequence bounded above converges to a finite real number. Its conclusion depends on the completeness of \(\mathbb{R}\). If we try to make the same claim for rational sequences, the conclusion can fail: a sequence can be increasing and bounded by a rational number while its real limit is irrational.
Proof. For \(n\in\mathbb{N}_0\), let $$ a_n=\frac{\lfloor 10^n\sqrt{7}\rfloor}{10^n}. $$ Each \(a_n\) is rational. To check that the sequence is nondecreasing, put \(x=10^n\sqrt{7}\). Since \(\lfloor x\rfloor\leq x\), we have \(10\lfloor x\rfloor\leq10x\), and therefore \(10\lfloor x\rfloor\leq\lfloor10x\rfloor\). Dividing by \(10^{n+1}\) gives $$ a_n=\frac{10\lfloor10^n\sqrt{7}\rfloor}{10^{n+1}} \leq\frac{\lfloor10^{n+1}\sqrt{7}\rfloor}{10^{n+1}} =a_{n+1}. $$ Also, the floor estimate gives \(a_n\leq\sqrt{7}<3\), so \(3\) is a rational upper bound.
The same estimate gives \(0\leq\sqrt{7}-a_n<10^{-n}\), so \(a_n\to\sqrt{7}\) in \(\mathbb{R}\). This limit is irrational: if \(\sqrt{7}=p/r\) in lowest terms, then \(p^2=7r^2\). Since \(7\) is prime and divides \(p^2\), it divides \(p\). Writing \(p=7k\) yields \(r^2=7k^2\), so \(7\) divides \(r\) as well, a contradiction. If \((a_n)\) converged to a rational number, uniqueness of real limits would force that number to be \(\sqrt{7}\). Thus it does not converge in \(\mathbb{Q}\), despite being nondecreasing and bounded above there. \(\square\)
The theorem is not a counterexample to the Monotone Convergence Theorem: that theorem concerns real sequences and supplies a real limit. Instead, it shows why completeness is needed when the theorem is applied within a particular number system. In \(\mathbb{Q}\), monotonicity and a rational upper bound do not ensure a rational limit.
Worked Example: A Bounded Set of Rationals with No Rational Supremum
Consider $$ S=\{q\in\mathbb{Q}:q>0\text{ and }q^2<3\}. $$ The set is nonempty because \(1\in S\), and it is bounded above by \(2\): if \(q\in S\) and \(q\geq2\), then \(q^2\geq4\), contradicting \(q^2<3\). In \(\mathbb{R}\), the least upper bound is \(\sqrt{3}\), which is not rational. We can verify directly that \(S\) has no supremum in \(\mathbb{Q}\).
Suppose, to the contrary, that \(s\in\mathbb{Q}\) is the supremum of \(S\) among rational numbers. Since \(1\in S\), \(s\geq1\). If \(s^2<3\), choose a positive rational \(h\) small enough that \(2sh+h^2<3-s^2\). Such an \(h\) exists by taking \(h=1/k\) for a sufficiently large positive integer \(k\). Then \((s+h)^2=s^2+2sh+h^2<3\), so \(s+h\in S\), contradicting that \(s\) is an upper bound.
If \(s^2>3\), choose a positive rational \(h<s\) small enough that \((s-h)^2>3\). For every \(q\in S\), we must have \(q<s-h\): otherwise \(q\geq s-h>0\) would imply \(q^2\geq(s-h)^2>3\), contrary to \(q^2<3\). Thus \(s-h\) is a rational upper bound for \(S\) that is smaller than \(s\), contradicting that \(s\) is the least rational upper bound. The only remaining possibility is \(s^2=3\), which would make \(s=\sqrt{3}\) rational, already shown to be impossible. Therefore \(S\) has no supremum in \(\mathbb{Q}\).
This example exposes another role of completeness. The real numbers have least upper bounds for nonempty sets bounded above; the rationals do not. In \(\mathbb{R}\), the supremum of \(S\) is exactly the number that the rational elements of \(S\) approach but never reach.
Incompleteness Does Not Mean Nothing Converges
It would be a mistake to conclude that a sequence of rational numbers cannot converge to a rational limit. Incompleteness says that some Cauchy sequences in the set fail to have limits in the set, not that all of them do. Many familiar rational sequences converge to rational numbers, and they can be handled using the same estimates as in \(\mathbb{R}\).
Worked Example: A Rational Cauchy Sequence That Does Converge in \(\mathbb{Q}\)
Let \(b_n=1-2^{-(n+1)}\) for \(n\in\mathbb{N}_0\). Each term is rational, and \(b_n\to1\), a rational limit. Directly, if \(m,n\geq N\), then $$ |b_m-b_n| =\big|2^{-(n+1)}-2^{-(m+1)}\big| \leq 2^{-(n+1)}+2^{-(m+1)} \leq 2^{-N}. $$ Given \(\varepsilon>0\), choose \(N\) such that \(2^{-N}<\varepsilon\). This proves that \((b_n)\) is Cauchy. Moreover, \(|b_n-1|=2^{-(n+1)}\to0\), so its limit is \(1\in\mathbb{Q}\). The contrast with the truncations of \(\sqrt{3}\) is not whether the terms are rational or whether the sequence is Cauchy; it is whether the real limit belongs to the set.
Where the Missing Limit Matters
Completeness is essential whenever an argument constructs approximations and then needs an actual object as their limit. Decimal truncations, nested approximations, and iterative procedures can produce terms that become arbitrarily close to one another. The Cauchy property records that increasingly accurate agreement. Completeness turns that agreement into a limit inside the space under consideration.
The real numbers support both the Cauchy Criterion and the Monotone Convergence Theorem because real Cauchy sequences have real limits and bounded monotone real sequences have real limits. In a smaller space, one must check that the limit stays in that space. The characterization from the previous tutorial provides a useful test: a nonempty subset of \(\mathbb{R}\) is complete exactly when it is closed in \(\mathbb{R}\). Since \(\mathbb{Q}\) is not closed in \(\mathbb{R}\), it is not complete.
The practical lesson is to identify which conclusion a proof actually needs. A proof that a sequence is Cauchy establishes closeness among late terms. It does not, by itself, identify a limit in an incomplete space. A proof that a sequence is bounded and monotone likewise needs an appropriate convergence theorem for the space being used. Whenever a limit is needed, completeness is the condition that prevents the desired object from disappearing outside the space.
Check Your Understanding
Use the examples and results above to explain where completeness enters each conclusion.
- Why does the sequence of rational decimal truncations of \(\sqrt{3}\) not contradict the Cauchy Criterion for Real Sequences?
- In the proof for the sequence approximating \(\sqrt{7}\), how is the inequality \(a_n\leq a_{n+1}\) obtained from the floor function?
- Why does the set \(\{q\in\mathbb{Q}:q>0,\ q^2<3\}\) have no rational supremum?
- Does incompleteness imply that every Cauchy sequence of rational numbers fails to converge in \(\mathbb{Q}\)? Give a reason for your answer.
- What extra fact, beyond being Cauchy, does completeness provide for a sequence in a set?